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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    If \[2x={{y}^{\frac{1}{5}}}+{{y}^{-\frac{1}{5}}}\] and \[({{x}^{2}}-1)\,\frac{{{d}^{2}}y}{d{{x}^{2}}}+\lambda x\,\frac{dy}{dx}+ky=0,\] then \[\lambda +k\] is equal to:              [JEE Online 09-04-2017]

    A)  26                                         

    B)  -24

    C)  -23                                        

    D)  -26

    Correct Answer: B

    Solution :

      \[{{y}^{1/5}}+{{y}^{-1/5}}=2x\] \[\left( \frac{1}{5}\,{{y}^{-4/5}}\,-\frac{1}{5}\,{{y}^{-6/5}} \right)\,\frac{dy}{dx}=2\] \[y'({{y}^{1/5}}\,-{{y}^{-1/5}})=10y\] \[y'\left( 2\sqrt{{{x}^{2}}-1} \right)\,=10y\] \[y''\left( 2\sqrt{{{x}^{2}}-1} \right)\,+y'2\frac{2x}{2\sqrt{{{x}^{2}}-1}}\,\sqrt{y'}\] \[y''({{x}^{2}}-1)\,+xy'\,=5\sqrt{{{x}^{2}}-1}(y')\] \[\] \[\lambda =1,\,\,k=-25\]


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